Problem 5 · AMC 8 Stretch
Core
Arithmetic & Operations
Algebra & Patterns
solve-an-easier-related-problemscale-the-answer
Find the sum \(7+77+777+7777+77777\) (five terms). Then describe the pattern for the sum of any number of such terms.
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Answer: 86415
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Hint 1 of 4
The 7's are awkward. Try the easier cousin first: \(9+99+999+\cdots\). A string of \(k\) nines is just \(10^k-1\) (for example \(999=1000-1\)).
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Hint 2 of 4
So \(9+99+999+9999+99999=(10-1)+(100-1)+(1000-1)+(10000-1)+(100000-1)\). Add the round numbers, then subtract the five 1's.
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Hint 3 of 4
Every digit 7 is exactly \(\tfrac79\) of a digit 9. So your 7's-sum is \(\tfrac79\) of the 9's-sum.
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Approach: Do the 9's first, then take 7/9
- Start with the easier version using 9's, because a block of \(k\) nines equals \(10^k-1\): \(9+99+999+9999+99999=(10+100+1000+10000+100000)-5\).
- The round numbers add to \(111110\), so the nines-sum is \(111110-5=111105\).
- Now scale by \(\tfrac79\), since every 7 is \(\tfrac79\) of a 9: \(7+77+777+7777+77777=\tfrac{7}{9}\times111105=7\times12345=86415\).
- Direct check by adding: \(7+77=84\), \(+777=861\), \(+7777=8638\), \(+77777=86415\).
- Pattern: the sum of the first \(n\) such terms is \(\dfrac{7\,(10^{n+1}-9n-10)}{81}\); a neat fact is that for \(n\le 9\) the answer is \(7\) times \(123\ldots n\) (here \(7\times12345=86415\)).
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